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Multiplication of a matrix by a scalar

by , PhD

This lecture explains how to multiply a matrix by a scalar.

Table of Contents

Definition

Remember that a scalar is just a single number, that is, a matrix having dimension $1 imes 1$.

Definition Let A be a $K imes L$ matrix and $lpha $ be a scalar. The product of A by $lpha $ is another $K imes L$ matrix, denoted by $lpha A$, such that its $left( k,l ight) $-th entry is equal to the product of $lpha $ by the $left( k,l ight) $-th entry of A, that is[eq1]for $1leq kleq K$ and $1leq lleq L$.

The product $Alpha $ could be defined in the same manner. However, the order of the product does not really matter, because $aA_{kl}=A_{kl}a$. Therefore, $Alpha $ can be considered the same as $lpha A$.

Example Let $lpha =2$ and define the $2 imes 3$ matrix[eq2]The product $lpha A$ is[eq3]

Properties

Basically all the properties enjoyed by multiplication of real numbers are inherited by multiplication of a matrix by a scalar.

Proposition (associative property) Multiplication of a matrix by a scalar is associative, that is,[eq4]for any matrix A and any scalars $lpha $ and $eta $.

Proof

Let A be a $K imes L$ matrix. We know that $eta A$ is another $K imes L $ matrix, such that its $left( k,l ight) $-th entry is equal to the product of $eta $ by the $left( k,l ight) $-th entry of A, that is,[eq5]Furthermore, [eq6] is a $K imes L$ matrix, such that its $left( k,l ight) $-th entry is equal to the product of $lpha $ by the $left( k,l ight) $-th entry of $eta A$, that is,[eq7]As a consequence, we have that[eq8]where we have used the definitions of [eq6] in step $ rame{A}$, $eta A$ in step $ rame{B}$ and [eq10] in step $ rame{D}$. In step $ rame{C}$ we have used the associativity of ordinary multiplication. Thus, we have proved that the $left( k,l ight) $-th entry of [eq6] is equal to the $left( k,l ight) $-th entry of [eq12]. Because this is true for every k and $l$, the statement is proved.

Proposition (distributive property 1) Multiplication of a matrix by a scalar is distributive with respect to matrix addition, that is,[eq13]for any scalar $lpha $ and any matrices A and $B$ such that their addition is meaningfully defined.

Proof

Let A and $B$ be $K imes L$ matrices. By the definition of matrix addition $A+B$ is another $K imes L$ matrix, such that its $left( k,l ight) $-th entry is equal to the sum of the $left( k,l ight) $-th entry of A and the $left( k,l ight) $-th entry of $B$, that is,[eq14]Furthermore, [eq15] is a $K imes L$ matrix, such that its $left( k,l ight) $-th entry is equal to the product of $lpha $ by the $left( k,l ight) $-th entry of $A+B$, that is,[eq16]As a consequence, we have that[eq17]where we have used the definitions of [eq15] in step $ rame{A}$, $A+B$ in step $ rame{B}$, $lpha A$ and $lpha B$ in step $ rame{D}$ and $lpha A+lpha B$ in step $ rame{E}$. In step $ rame{C}$ we have used the distributivity of ordinary multiplication. Thus, we have proved that the $left( k,l ight) $-th entry of [eq15] is equal to the $left( k,l ight) $-th entry of $lpha A+lpha B$. Because this is true for every k and $l$, the statement is proved.

Proposition (distributive property 2) Multiplication of a matrix by a scalar is distributive with respect to the addition of scalars, that is,[eq20]for any scalars $lpha $ and $eta $ and any matrix A.

Proof

Let A be a $K imes L$ matrix. We know that [eq21] is another $K imes L$ matrix, such that its $left( k,l ight) $-th entry is equal to the product of $lpha +eta $ by the $left( k,l ight) $-th entry of A, that is,[eq22]As a consequence, we have that[eq23]where we have used the definitions of [eq24] in step $ rame{A}$, $lpha A$ and $eta A$ in step $ rame{C}$, $lpha A+eta A$ in step $ rame{D}$. In step $ rame{B}$ we have used the distributivity of ordinary multiplication. Thus, we have proved that the $left( k,l ight) $-th entry of [eq24] is equal to the $left( k,l ight) $-th entry of $lpha A+eta A$. Because this is true for every k and $l$, the statement is proved.

Solved exercises

Below you can find some exercises with explained solutions.

Exercise 1

Let A be the following $3 imes 3$ matrix[eq26]Let $lpha =3$. Compute the product $lpha A$.

Solution

The product $lpha A$ is another $3 imes 3$ matrix such that for each $1leq kleq 3$ and $1leq lleq 3$, the $left( k,l ight) $-th element of $lpha A$ is equal to the product between $lpha $ and the $left( k,l ight) $-th element of A: [eq27]

Exercise 2

Let A be a $1 imes 2$ row vector defined by[eq28]and $B$ a $2 imes 2$ matrix defined by[eq29]Compute the product[eq30]where $A^{ op }$ denotes the transpose of A.

Solution

The transpose of A is[eq31]The product between A and its transpose is[eq32]which is a scalar. As a consequence, we have that[eq33]

Exercise 3

Define two $1 imes 2$ row vectors:[eq34]Find a scalar $lpha $ such that[eq35]where[eq36]

Solution

By applying the definition of multiplication of a matrix by a scalar, we obtain[eq37]By applying the definition of matrix addition, we get[eq38]Therefore, the equation[eq39]is satisfied if and only if[eq40]which in turn is satisfied if and only if[eq41]But this implies[eq42]

How to cite

Please cite as:

Taboga, Marco (2021). "Multiplication of a matrix by a scalar", Lectures on matrix algebra. https://www.statlect.com/matrix-algebra/multiplication-of-a-matrix-by-a-scalar.

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